标签：快速 queue 更新 首部 top nbsp tin 部分 sum
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
For example,
Given [3,2,1,5,6,4] and k = 2, return 5.
Note:
You may assume k is always valid, 1 ≤ k ≤ array‘s length.
方法一：利用快速排序的partition的部分，这样每次都能知道第一个数的左边有多少个数，与k比较大小，分类讨论。注意点，while中是小于号，注意与二分查找的不同，整个执行结束low==high。
class Solution { public: int helper(vector<int>& nums, int k, int begin, int end){ int temp=nums[begin]; int origBegin=begin,origEnd=end; while(begin<end){ while(begin<end&&nums[end]<=temp)end--; nums[begin]=nums[end]; while(begin<end&&nums[begin]>=temp)begin++; nums[end]=nums[begin]; } nums[begin]=temp; //cout<<begin<<‘ ‘<<end<<endl; if(begin-origBegin+1==k)return nums[begin]; else if(begin-origBegin+1<k)return helper(nums,k-begin-1+origBegin,begin+1,origEnd); else return helper(nums,k,origBegin,end); } int findKthLargest(vector<int>& nums, int k) { return helper(nums,k,0,nums.size()-1); } };
方法二：利用优先队列，先塞进去k个数，碰到比最小值更大的就更新，最后的队列首部肯定就是要求的第k大值。
class Solution { public: int findKthLargest(vector<int>& nums, int k) { priority_queue<int, vector<int>, greater<int>> minHeap; for(int i = 0; i<nums.size(); i++){ if(i<k) minHeap.push(nums[i]); else{ int topv = minHeap.top(); if(topv < nums[i]) {minHeap.pop(); minHeap.push(nums[i]);} } } return minHeap.top(); } };
方法三：利用stl的nth_element.
class Solution { public: int findKthLargest(vector<int>& nums, int k) { int n = nums.size(); nth_element(nums.begin(), nums.begin() + n - k, nums.end()); return nums[n - k]; } };
215. Kth Largest Element in an Array
标签：快速 queue 更新 首部 top nbsp tin 部分 sum
原文地址：http://www.cnblogs.com/tsunami-lj/p/7631864.html
